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#volume calculation

13 public questions tagged with this topic.

Partial Pressures and Gas Mixtures

What is the volume of 0.2 moles of an ideal gas at 1.5 atm and 227°C? (R = 8.31 J mol⁻¹ K⁻¹)

**Partial pressure concept** for mixture of non-reacting gases, total pressure sum of partials, each gas behaves independently, kinetic theory still holds with effective n = Σ n_i, mean free path depends on total n and cross-sections, mixture properties weighted by mole tions. PV = μ R T, V = (μ R T)/(P).T = 227 + 273 = 500 K, P = 1.5 × 1.01 × 10⁵ = 1.515 × 10⁵ Pa.V = (0.2 × 8.31 × 500)/(1.515 × 10⁵) = 5.485 × 10⁻³ m³ ≈ 5.49 litres. Substituting values gives 5.49 litres, which matches expected kinetic theory result, confirming mean free path λ = 1/(√2

Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > Partial Pressures and Gas Mixtures

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Molecular Mass Density and Ideal Gas Equation

A gas at 1 atm and 273 K has a volume of 15 litres. If the temperature increases to 819 K at constant pressure, what is

**Molecular mass and density** relation ρ = P M/(R T) allows density calculation, ideal gas law also P = n k_B T where n number density, molecular mass determines mass per molecule, density increases with pressure and decreases with temperature, inverse T dependence. Charles’ law: (V₁)/(T₁) = (V₂)/(T₂).V₁ = 15 litres, T₁ = 273 K, T₂ = 819 K.V₂ = V₁ × (T₂)/(T₁) = 15 × (819)/(273) = 45 litres. Substituting values gives 45 litres, which matches expected kinetic theory result, confirming mean free path λ = 1/(√2 n π d²), ideal gas law P V = n R T and v_rms = √(3 R T/M) relations.

Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > Molecular Mass Density and Ideal Gas Equation

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RMS Speed and Temperature Dependence

What is the volume of 0.6 moles of an ideal gas at 3 atm and 527°C? (R = 8.31 J mol⁻¹ K⁻¹)

**Maxwell-Boltzmann distribution** gives distribution of speeds, v_rms = √(3kT/m), most probable v_mp = √(2kT/m), average v_avg = √(8kT/πm), all ∝ √T, ratio v_rms:v_avg:v_mp =1.732:1.596:1.414, temperature raises all speeds proportionally. PV = μ R T, V = (μ R T)/(P).T = 527 + 273 = 800 K, P = 3 × 1.01 × 10⁵ = 3.03 × 10⁵ Pa.V = (0.6 × 8.31 × 800)/(3.03 × 10⁵) = 1.317 × 10⁻² m³ ≈ 13.17 litres. Substituting values gives 13.2 litres, which matches expected kinetic theory result, confirming mean free path λ = 1/(√2 n π d²), ideal gas law P V = n R T and

Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > RMS Speed and Temperature Dependence

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Internal Energy of Ideal Gases

A gas at 1.5 atm and 300 K has a volume of 12 litres. If the temperature increases to 450 K at constant pressure, what i

**Equipartition theorem** energy ½ k_B T per degree of freedom per molecule, f degrees give U = f/2 k_B T per molecule, f/2 R T per mole, internal energy function of T only for ideal gas, no intermolecular potential. Charles’ law: (V₁)/(T₁) = (V₂)/(T₂).V₁ = 12 litres, T₁ = 300 K, T₂ = 450 K.V₂ = V₁ × (T₂)/(T₁) = 12 × (450)/(300) = 18 litres. Substituting values gives 18 litres, which matches expected kinetic theory result, confirming mean free path λ = 1/(√2 n π d²), ideal gas law P V = n R T and v_rms = √(3 R T/M) relations.

Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > Internal Energy of Ideal Gases

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Gas Laws and Volume-Temperature Relations

A gas at 1 atm and 273 K has a volume of 8 litres. If the pressure increases to 4 atm at constant temperature, what is t

**Gas laws** Boyle, Charles, Gay-Lussac are special cases of ideal gas equation, for constant pressure volume-temperature relation V ∝ T, for constant temperature pressure-volume inverse, for constant volume pressure-temperature direct, enabling calculation of new volume from temperature ratio. Boyle’s law: P₁ V₁ = P₂ V₂.P₁ = 1 atm, V₁ = 8 litres, P₂ = 4 atm.V₂ = (P₁ V₁)/(P₂) = (1 × 8)/(4) = 2 litres. Substituting values gives 2 litres, which matches expected kinetic theory result, confirming mean free path λ = 1/(√2 n π d²), ideal gas law P V = n R T and v_rms = √(3 R T/M) relations.

Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > Gas Laws and Volume-Temperature Relations

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Gas Laws and Volume-Temperature Relations

A gas at 2 atm and 400 K has a volume of 10 litres. If the temperature decreases to 100 K at constant pressure, what is

**Charles' law** V₁/T₁ = V₂/T₂ at constant pressure, volume proportional to absolute temperature (K), Gay-Lussac P₁/T₁ = P₂/T₂ at constant volume, Boyle's law P₁V₁ = P₂V₂ at constant temperature, combined ideal gas law P V = n R T, R=8.314 J/mol·K. For V₁=24 L T₁=300 K T₂=600 K, V₂= V₁ T₂/T₁=48 L, volume doubles when T doubles at constant P. Charles’ law: (V₁)/(T₁) = (V₂)/(T₂).V₁ = 10 litres, T₁ = 400 K, T₂ = 100 K.V₂ = V₁ × (T₂)/(T₁) = 10 × (100)/(400) = 2.5 litres. Substituting values gives 2.5 litres, which matches expected kinetic theory result, confirming mean free path λ = 1/(√2

Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > Gas Laws and Volume-Temperature Relations

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Gas Laws and Volume-Temperature Relations

A gas at 3 atm and 400 K has a volume of 15 litres. If the pressure drops to 1.5 atm at constant temperature, what is th

**Charles' law** V₁/T₁ = V₂/T₂ at constant pressure, volume proportional to absolute temperature (K), Gay-Lussac P₁/T₁ = P₂/T₂ at constant volume, Boyle's law P₁V₁ = P₂V₂ at constant temperature, combined ideal gas law P V = n R T, R=8.314 J/mol·K. For V₁=24 L T₁=300 K T₂=600 K, V₂= V₁ T₂/T₁=48 L, volume doubles when T doubles at constant P. Boyle’s law: P₁ V₁ = P₂ V₂.P₁ = 3 atm, V₁ = 15 litres, P₂ = 1.5 atm.V₂ = (P₁ V₁)/(P₂) = (3 × 15)/(1.5) = 30 litres. Substituting values gives 30 litres, which matches expected kinetic theory result, confirming mean free path λ =

Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > Gas Laws and Volume-Temperature Relations

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PHYSICS

What is the volume of a nucleus with radius 3.0 × 10⁻¹⁵ m ? (Use π = 3.14 )

Given: What is the volume of a nucleus with radius 3.0 × 10⁻¹⁵ m ? (Use π = 3.14 ) These values define the system as per NCERT data. Formula: Volume = 4/3 π R³. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: R³ = (3.0 × 10⁻¹⁵)³ = 2.7 × 10⁻⁴⁴ m³ . Volume = 4/3 × 3.14 × 2.7 × 10⁻⁴⁴ approx 1.13 × 10⁻⁴³ m³ . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Dual Nature, Atoms, Nuclei and Electronic Devices, Topic: Atomic structure, nuclear binding and semiconductor physics.

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